Answer:
a) To find the first five terms of the linear sequence and the sum of the first n terms, we can use the following formulas:
The common difference of an A.P is (an - a1) / (n - 1)
The sum of the first n terms of an A.P is (n/2)(2a1 + (n-1)d)
Let the first term of the linear sequence be a1 and the common difference be d. Since the first and second terms of the exponential sequence are the first and third terms of the linear sequence, we can write:
a1 = a and a3 = ar^2
Given that the fourth term of the linear sequence is 10, we can write:
a1 + 3d = 10
Given that the sum of the first five terms of the linear sequence is 60, we can write:
(5/2)(2a1 + (5-1)d) = 60
Solving these two equations for a1 and d gives:
a1 = 2, d = 2
Therefore, the first five terms of the linear sequence are 2, 4, 6, 8, 10, and the sum of the first n terms is (n/2)(2a1 + (n-1)d) = (n/2)(2*2 + (n-1)2) = n2
b) To find the sum of the first n terms of the exponential sequence, we can use the following formula:
Sn = a(1 - r^n) / (1 - r)
where a is the first term and r is the common ratio.
Given that a1 = a and a3 = ar^2, we can write:
a = a1 and r^2 = a3/a1
Substituting these values into the formula for Sn gives:
Sn = a(1 - r^n) / (1 - r) = a1(1 - (a3/a1)^n) / (1 - (a3/a1)^(1/2))
c) As n tends to infinity, the limit of Sn is a1/(1-(a3/a1)^(1/2))
So the first five terms of the linear sequence are 2, 4, 6, 8, 10, the sum of the first n terms is 2n and the limit of Sn for large value of n is a1/(1-(a3/a1)^(1/2))
Explanation: